If vector ` vec a = hati + hatj + hatk , vecb = 4 hati + 3 hatj + 4 hatk ` and ` vec c = hati + alpha hatj + beta hatk ` are linearly dependent and ` | vec c | = sqrt3 ` , then value of ` | alpha | + | beta | ` is

Options

A. 2
B. 3
C. 1
D. 4

Explanation

To solve the problem, we need to determine the values of \(|\alpha|\) and \(|\beta|\) given that the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are linearly dependent and that \(|\vec{c}| = \sqrt{3}\).

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = 4\hat{i} + 3\hat{j} + 4\hat{k}, \quad \vec{c} = \hat{i} + \alpha \hat{j} + \beta \hat{k} \]

2. **Condition for linear dependence**:
The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are linearly dependent if the determinant of the matrix formed by their coefficients is zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & \alpha \\ 4 & 4 & \beta \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ 1 \cdot (3\beta - 4\alpha) - 1 \cdot (4\beta - 4) + 1 \cdot (4\alpha - 3) = 0 \]
Simplifying this gives:
\[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \]
Combining like terms:
\[ (3\beta - 4\beta) + (-4\alpha + 4\alpha) + (4 - 3) = 0 \implies -\beta + 1 = 0 \]
Thus, we find:
\[ \beta = 1 \]

4. **Magnitude of vector \(\vec{c}\)**:
We know that \(|\vec{c}| = \sqrt{3}\):
\[ |\vec{c}| = \sqrt{1 + \alpha^2 + \beta^2} = \sqrt{3} \]
Squaring both sides:
\[ 1 + \alpha^2 + \beta^2 = 3 \]
Substituting \(\beta = 1\):
\[ 1 + \alpha^2 + 1^2 = 3 \implies 1 + \alpha^2 + 1 = 3 \implies \alpha^2 + 2 = 3 \]
Thus:
\[ \alpha^2 = 1 \implies \alpha = \pm 1 \]

5. **Calculate \(|\alpha| + |\beta|\)**:
Now we have:
\[ |\alpha| = 1 \quad \text{and} \quad |\beta| = 1 \]
Therefore:
\[ |\alpha| + |\beta| = 1 + 1 = 2 \]

### Final Answer:
\[ |\alpha| + |\beta| = 2 \]